# How do you find the average rate of change of f(x)=x^2+2 over [0,2]?

Mar 19, 2016

By calculus: average rate of change is 2

#### Explanation:

As you have used $f \left(x\right)$ I am assuming you are using Calculus.

$\textcolor{b l u e}{\text{Short cut method}}$

By sight: $\text{rate of change } \to \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

At $x = 0 \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(0\right) = 0$

At $x = 2 \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(2\right) = 4$

Thus the average rate of change is

$\frac{0 + 4}{2} = 2$
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$\textcolor{b l u e}{\text{From first principles}}$

Let $\text{ } y = {x}^{2} + 2$..........................(1)

Increment $x \text{ by the very small amount of } \delta x$

As $x$ has change then $y$ will change as well

Let the change in $y \text{ be } \delta y$

Now we have

$y + \delta y = {\left(x + \delta x\right)}^{2} + 2$

$y + \delta y = {x}^{2} + 2 x \delta x + 2$.....................(2)

Subtract equation (1) from equation (2)

$y + \delta y = {x}^{2} + 2 x \delta x + 2$
$\underline{y \text{ "=x^2" } + 2}$ apply the subtraction
$\text{ "deltay = 0} + 2 x \delta x + 0$

Divide by $\delta x$

$\frac{\delta y}{\delta x} = 2 x \times \frac{\delta x}{\delta x}$

but $\frac{\delta x}{\delta x} = 1$

So

${\lim}_{\delta x \to 0} \frac{\delta y}{\delta x} = \frac{\mathrm{dx}}{\mathrm{dy}} = 2 x$

Then solve for average rate of change as above.

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Note that${\lim}_{\delta x \to 0}$ means consider the situation where

$\delta x$ gets so small it may as well be zero, but in reality it has not quite got there!

Mar 19, 2016

2

#### Explanation:

Average rate = $\frac{\int \frac{d}{\mathrm{dx}} f \left(x\right) \mathrm{dx}}{\int \mathrm{dx}}$, between the limits.
= $\frac{\int \mathrm{df} \left(x\right)}{\int \mathrm{dx}}$, between the limits.
= (f(2)-f(0))/((2-0) = $\frac{6 - 2}{2}$.